Calculating Refraction Angles of Light Through a Prism | Snell's Law | n=1.46

  • Thread starter Thread starter cmilho10
  • Start date Start date
  • Tags Tags
    Light Prism
cmilho10
Messages
20
Reaction score
0
Light of wavelength 700 nm is incident on the face of a fused quartz prism at an angle of 80.0° (with respect to the normal to the surface). The apex angle of the prism is 60.0°.

Use the value of n from Figure 35.20, to calculate the following angles.
(a) the angle of refraction at the first surface
(b) the angle of incidence at the second surface
(c) the angle of refraction at the second surface
(d) the angle between the incident and emerging rays

n=1.46

I used snell's law in order to find (a) which is 42 degrees, but when i try to do the geometry for the other parts it says my answers are wrong (i got 18 degrees for part b). It is hard to explain without drawing a picture exactly where i went wrong, but if anybody has any suggestions i would appreciate it.
 
Physics news on Phys.org
cmilho10 said:
Light of wavelength 700 nm is incident on the face of a fused quartz prism at an angle of 80.0° (with respect to the normal to the surface). The apex angle of the prism is 60.0°.
Use the value of n from Figure 35.20, to calculate the following angles.
(a) the angle of refraction at the first surface
(b) the angle of incidence at the second surface
(c) the angle of refraction at the second surface
(d) the angle between the incident and emerging rays
n=1.46
I used snell's law in order to find (a) which is 42 degrees, but when i try to do the geometry for the other parts it says my answers are wrong (i got 18 degrees for part b).

I have to say I don't find any error. Using Snell's law you have indeed 42.4 degrees for (a), and if the apex angle is 60 degrees, this means that you have 17.6 degrees (60 - 42.4) on the other side for (b) wrt to the other normal...
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...

Similar threads

Replies
1
Views
2K
Replies
9
Views
2K
Replies
207
Views
12K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
4
Views
3K
Replies
2
Views
2K
Back
Top